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趣题: 证明等式

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趣题: 证明等式 #

题目 #

给定整数 $a, b$ , 有 $a \leq b$ . 证明: $$ \sum_{i = 0}^{a-1} \frac{a(a-1)…(a-i)}{b(b-1)…(b-i)} = \frac{a}{b-a+1} $$

Solution 1 #

记 $f(x,y) = \sum_{i = 0}^{x-1} \frac{x(x-1)…(x-i)}{y(y-1)…(y-i)} - \frac{x}{y-x +1}$ . 如果有 $f(a, b) = 0$ , 那么 $$ f(a +1,b+1) =\sum_{i = 0}^{a} \frac{(a+1)\cdot a…(a + 1 -i)}{(b+1)\cdot b…(b + 1-i)} - \frac{a+1}{(b+ 1) - (a + 1) + 1}\= \frac{a+1}{b+1}(1 + \sum_{i = 0}^{a-1} \frac{a(a-1)…(a-i)}{b(b-1)…(b-i)}) - \frac{a+1}{(b+ 1) - (a + 1) + 1} \= \frac{a+1}{b+1}(1 + \frac{a}{b-a+1}) - \frac{a+1}{(b+ 1) - (a + 1) + 1} \= \frac{a+1}{(b+ 1) - (a + 1) + 1} - \frac{a+1}{(b+ 1) - (a + 1) + 1} \=0 $$ 对于 $f(x, y), x\leq y$ , 考虑 $$ f(1, y - x)=\frac{1}{y-x} - \frac{1}{y-x - 1 + 1}= 0 $$ 故 $$ f(x, y) = f(x-1, y - 1) = \dots =f(1, y -x)=0 $$ 所以原式成立, 证毕.